Saving Beans
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1666 Accepted Submission(s): 592
in n different trees. However, since the food is not sufficient nowadays, they will get no more than m beans. They want to know that how many ways there are to save no more than m beans (they are the same) in n trees.
Now they turn to you for help, you should give them the answer. The result may be extremely huge; you should output the result modulo p, because squirrels can’t recognize large numbers.
Then followed T lines, each line contains three integers n, m, p, means that squirrels will save no more than m same beans in n different trees, 1 <= n, m <= 1000000000, 1 < p < 100000 and p is guaranteed to be a prime.
2 1 2 5 2 1 5
3 3HintHint For sample 1, squirrels will put no more than 2 beans in one tree. Since trees are different, we can label them as 1, 2 … and so on. The 3 ways are: put no beans, put 1 bean in tree 1 and put 2 beans in tree 1. For sample 2, the 3 ways are: put no beans, put 1 bean in tree 1 and put 1 bean in tree 2.
一道很强悍的数论题目,推都推不出来,但是好在,有别人的思路,明天再推推,看看能不能推出来,
但是有很多的知识点,比如说,费马小定理,比如说,lucas还比如说求c(n,m)
这些都是很基础的算法,但是自己以前都不知道,这一道题目都让我全部用到了,自己果然是菜鸟中的菜鸟,不过学到了知识,自己还是很开心的.
拿来别人的解题报告看看.
然后贴出自己的代码
题目相当于求n个数的和不超过m的方案数。
如果和恰好等于m,那么就等价于方程x1+x2+...+xn = m的解的个数,利用插板法可以得到方案数为:
(m+1)*(m+2)...(m+n-1) = C(m+n-1,n-1) = C(m+n-1,m)
现在就需要求不大于m的,相当于对i = 0,1...,m对C(n+i-1,i)求和,根据公式C(n,k) = C(n-1,k)+C(n-1,k-1)得
C(n-1,0)+C(n,1)+...+C(n+m-1,m)
= C(n,0)+C(n,1)+C(n+1,2)+...+C(n+m-1,m)
= C(n+m,m)
现在就是要求C(n+m,m) % p,其中p是素数。
然后利用Lucas定理的模板就可以轻松的求得C(n+m,m) % p的值
下面简单介绍一下Lucas定理:
Lucas定理是用来求 C(n,m) mod p的值,p是素数(从n取m组合,模上p)。
描述为:
Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)
Lucas(x,0,p)=1;
简单的理解就是:
以求解n! % p 为例,把n分段,每p个一段,每一段求得结果是一样的。但是需要单独处理每一段的末尾p,2p,...,把p提取出来,会发现剩下的数正好又是(n/p)! ,相当于
划归了一个子问题,这样递归求解即可。
这个是单独处理n!的情况,当然C(n,m)就是n!/(m! *(n-m)!),每一个阶乘都用上面的方法处理的话,就是Lucas定理了
Lucas最大的数据处理能力是p在10^5左右。
而C(a,b) =a! / ( b! * (a-b)! ) mod p
其实就是求 ( a! / (a-b)!) * ( b! )^(p-2) mod p
(上面这一步变换是根据费马小定理:假如p是质数,且a,p互质,那么a的(p-1)次方除以p的余数恒为1,
那么a和a^(p-2)互为乘法逆元,则(b / a) = (b * a^(p-2) ) mod p)
用下面的Lucas定理程序实现就能得出结果,实现过程中要注意乘法时的强制转换
#include <stdio.h> #include <string.h> #include <iostream> #include <string> using namespace std; int N, M, P; int fac[100011]; void factorial() { fac[0] = 1; for (int i = 1; i <= P; i++) { fac[i] = (long long int)fac[i - 1] * i % P; } } int quick_pow(int a, int n) { int ans = 1; while (n) { if (n & 1) { ans = (long long int)ans * a % P; } a = (long long int) a * a % P; n >>= 1; } return ans; } int C(int n, int m) { if (n < m) { return 0; } return (long long int)fac[n] * (quick_pow((long long int)fac[m] * fac[n - m] % P, P - 2)) % P; } int lucas(long long int n, long long int m) { if (m == 0) { return 1; } return (long long int)C(n % P, m % P) * lucas(n / P, m / P) % P; } int main() { int T; scanf("%d", &T); while (T--) { scanf("%d%d%d", &N, &M, &P); factorial(); int ans = lucas(N + M, M); printf("%d\n", ans); } // system("pause"); return 0; }